A) \[{{\operatorname{V}}^{2}}_{f}={{V}_{1}}^{2}+\frac{2GM}{{{M}_{e}}R}\left[ 1-\frac{1}{10} \right]\]
B) \[{{\operatorname{V}}^{2}}_{f}={{V}_{1}}^{2}+\frac{2GM}{{{R}_{e}}}\left[ 1+\frac{1}{10} \right]\]
C) \[{{\operatorname{V}}^{2}}_{f}={{V}_{1}}^{2}+\frac{2GMe}{{{R}_{e}}}\left[ 1-\frac{1}{10} \right]\]
D) \[{{\operatorname{V}}^{2}}_{f}={{V}_{1}}^{2}+\frac{2GM}{{{R}_{e}}}\left[ 1-\frac{1}{10} \right]\]
Correct Answer: C
Solution :
\[\operatorname{Kinetic} energy of asteroid =\frac{1}{2}m{{v}^{2}}\] \[\operatorname{P}.E of asteroid =\frac{-GMm}{10{{R}_{e}}}\] Apply law of energy conservation \[{{\operatorname{K}}_{1}}+{{P}_{1}}={{K}_{2}}+{{P}_{2}}\] \[\frac{1}{2}m{{v}_{1}}^{2}-\frac{GMm}{10{{R}_{e}}}+\frac{1}{2}{{\operatorname{mv}}_{\operatorname{f}}}^{2}-\frac{GMm}{{{R}_{e}}}\] \[\frac{{{v}_{1}}^{2}}{2}-\frac{GM}{10{{R}_{e}}}+\frac{GM}{{{R}_{e}}}=\frac{{{v}_{\operatorname{f}}}^{2}}{2}\] \[{{v}_{\operatorname{f}}}^{2}={{v}_{1}}^{2}+\frac{2GM}{{{R}_{e}}}\left[ 1-\frac{1}{10} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
